Author 
Message 
gt7er

Post subject: is there a formula to calculate the number of factors? Posted: Sat Mar 08, 2008 1:35 pm 


what is the number of factors for 441? It breaks down into 3, 3, 7, 7 ...how do you find the total number of factors? what's the formula?





H_to_750

Post subject: no formula Posted: Sat Mar 08, 2008 3:39 pm 


There is no formula you just take a look at how many different combinations there are of prime factors
So in this case
WE have 3, 7, (7 x 7 = 49), (7 x 3 = 21), (3 x 3 = 9) (9 x 7 = 63), (49 x 3 = 147) , 441, and 1
So for 441 we have total of 9 factors. Just always remember that 1 and the number it self is also a factor.





David Pollack

Post subject: A nice formula for the number of factors Posted: Sun Mar 09, 2008 1:25 am 


Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.
Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.
So the number of factors for your number would be (a+1)*(b+1)*...(k+1).
For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.
One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24.





RonPurewal

Post subject: Re: A nice formula for the number of factors Posted: Wed Mar 12, 2008 3:59 am 


ManhattanGMAT Staff 

Posts: 12726

David Pollack wrote: Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.
Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.
So the number of factors for your number would be (a+1)*(b+1)*...(k+1).
For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.
One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24.
this is just awesome. very well done.
i have some unsolicited advice for you:
step 1) drop whatever you're doing
step 2) write a textbook
step 3) resume whatever you're doing
thank you for saving us the time required to answer this ourselves. :)





Guest

Post subject: Re: A nice formula for the number of factors Posted: Thu Mar 20, 2008 3:08 am 


RPurewal wrote: David Pollack wrote: Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.
Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.
So the number of factors for your number would be (a+1)*(b+1)*...(k+1).
For example, if your number is 441 = 3^2 * 7^2 then the number of factors is (2+1)*(2+1)=3*3=9.
One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24. this is just awesome. very well done. i have some unsolicited advice for you: step 1) drop whatever you're doing step 2) write a textbook step 3) resume whatever you're doing thank you for saving us the time required to answer this ourselves. :)
That explanation less the examples read more like directions to fix a pinball machine in chinese





Dlew 686

Post subject: Posted: Thu Mar 20, 2008 11:40 pm 






iillondon

Post subject: Prime factors ... quickest way ? Posted: Mon Mar 24, 2008 6:36 am 


Hi there ...
what is the quickest way to find the prime factors for a number ?
How would you find the prime factors of 3841, for example ?
Does anyone know of a quick way to do it.
Thanks for the help. Appreciate it.





upa

Post subject: Re: Prime factors ... quickest way ? Posted: Mon Mar 24, 2008 9:49 pm 


iillondon wrote: Hi there ...
what is the quickest way to find the prime factors for a number ? How would you find the prime factors of 3841, for example ? Does anyone know of a quick way to do it.
Thanks for the help. Appreciate it.
I guess there is not. If anybody has I appreciate him/her for his/her brilliant idea.
Just factorize it. I know it is not easy and doesnot have factors 2, 3, 5, 7, 11, 13, 17, and 19. but 23 divides it.
3841 = 23 x 167
i guess 167 is also a prime cuz if it were not a prime it has to be divided by 2, 3, 5, 7, or son. none of them divides 167, so its a prime.
this is a tough one.





StaceyKoprince

Post subject: Posted: Thu Mar 27, 2008 2:29 am 


ManhattanGMAT Staff 

Posts: 7637 Location: San Francisco

David  I agree, nice process. And for those of you who weren't quite sure how to follow the theory  don't worry about it! Just follow the examples :)
I don't know of a shortcut to find the complete set of prime factors for a number very quickly. I'm guessing there isn't one because one of the weird things about primes is that they pretty much defy our attempts to reduce things to typical patterns or formulas.
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT





Guest

Post subject: finding primes Posted: Wed Jan 21, 2009 9:10 am 


Not exactly a shortcut to finding the primes, but to help determine a large prime. Take the square root of the number. You only have to try the primes up through the one just larger than the square root.
In other words, sqrt 167 = 12.9... so you only have to try dividing it by the prime numbers through 13. Since none of those work, it is prime.





JonathanSchneider

Post subject: Re: is there a formula to calculate the number of factors? Posted: Fri Feb 13, 2009 3:49 pm 


ManhattanGMAT Staff 

Posts: 477 Location: Durham, NC

Good point  glad you added that!





zxtonizx

Post subject: Re: is there a formula to calculate the number of factors? Posted: Sun Jan 01, 2012 2:04 pm 


Course Students 

Posts: 1

Cannot tell you the time David Pollack's shortcut has saved. Thank you sir!





tim

Post subject: Re: is there a formula to calculate the number of factors? Posted: Tue Jan 10, 2012 7:28 pm 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

cool
_________________ Tim Sanders Manhattan GMAT Instructor





jp.jprasanna

Post subject: Re: is there a formula to calculate the number of factors? Posted: Tue May 15, 2012 12:11 pm 


Students 

Posts: 203

Hi  I undestand the below method, which gives me total factors, but is there a way I can find the numbers itself? Apart from listing them out!
One more example: if your number is 360 = 8*9*5 = 2^3 * 3^2 * 5^1 then the number of factors is (3+1)*(2+1)*(1+1)=24
Cheers





tim

Post subject: Re: is there a formula to calculate the number of factors? Posted: Tue May 22, 2012 3:43 am 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

no, you just have to list them out..
_________________ Tim Sanders Manhattan GMAT Instructor





